8.2: Applications of the Pythagorean Theorem
What if you had a 52" High Definition Television (52" being the length of the diagonal of the rectangular viewing area)? High Definition Televisions (HDTVs) have sides in the ratio of 16:9. What is the length and width of a 52” HDTV? What is the length and width of an HDTV with a \begin{align*}y''\end{align*}
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CK-12 Foundation: Chapter8ApplicationsofthePythagoreanTheoremA
James Sousa: Pythagorean Theorem and Its Converse
Guidance
There are many different applications of the Pythagorean Theorem. Three applications are explored below.
Find the Height of an Isosceles Triangle
One way to use The Pythagorean Theorem is to identify the heights in isosceles triangles so you can calculate the area. The area of a triangle is \begin{align*}\frac{1}{2} \ bh\end{align*}
If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height.
Prove the Distance Formula
Another application of the Pythagorean Theorem is the Distance Formula.
First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to find these distances.
Now that we have a right triangle, we can use the Pythagorean Theorem to find \begin{align*}d\end{align*}
Distance Formula: The distance \begin{align*}A(x_1, y_1)\end{align*}
Determine if a Triangle is Acute, Obtuse, or Right
We can extend the converse of the Pythagorean Theorem to determine if a triangle has an obtuse angle or is acute. We know that if the sum of the squares of the two smaller sides equals the square of the larger side, then the triangle is right. We can also interpret the outcome if the sum of the squares of the smaller sides does not equal the square of the third.
Theorem: (1) If the sum of the squares of the two shorter sides in a right triangle is greater than the square of the longest side, then the triangle is acute. (2) If the sum of the squares of the two shorter sides in a right triangle is less than the square of the longest side, then the triangle is obtuse.
In other words: The sides of a triangle are \begin{align*}a, b\end{align*}
If \begin{align*}a^2 + b^2 > c^2\end{align*}
If \begin{align*}a^2 + b^2 = c^2\end{align*}
If \begin{align*}a^2 + b^2 < c^2\end{align*}
Proof of Part 1:
Given: In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*}
In \begin{align*}\triangle LMN, \angle N\end{align*}
Prove: \begin{align*}\triangle ABC\end{align*}
Statement | Reason |
---|---|
1. In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*} |
Given |
2. \begin{align*}a^2 + b^2 = h^2\end{align*} |
Pythagorean Theorem |
3. \begin{align*}c^2 < h^2\end{align*} |
Transitive PoE |
4. \begin{align*}c < h\end{align*} |
Take the square root of both sides |
5. \begin{align*}\angle C\end{align*} |
The largest angle is opposite the longest side. |
6. \begin{align*}m \angle N = 90^\circ\end{align*} |
Definition of a right angle |
7. \begin{align*}m \angle C < m \angle N\end{align*} |
SSS Inequality Theorem |
8. \begin{align*}m \angle C < 90^\circ\end{align*} |
Transitive PoE |
9. \begin{align*}\angle C\end{align*} |
Definition of an acute angle |
10. \begin{align*}\triangle ABC\end{align*} |
If the largest angle is less than \begin{align*}90^\circ\end{align*} |
Example A
What is the area of the isosceles triangle?
First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, find the length of the altitude using the Pythagorean Theorem.
\begin{align*}7^2 + h^2 & = 9^2\\ 49 + h^2 & = 81\\ h^2 & = 32\\ h & = \sqrt{32} = 4 \sqrt{2}\end{align*}
Now, use \begin{align*}h\end{align*} and \begin{align*}b\end{align*} in the formula for the area of a triangle.
\begin{align*}A = \frac{1}{2} \ bh = \frac{1}{2} (14) \left (4 \sqrt{2} \right ) = 28 \sqrt{2} \ units^2\end{align*}
Example B
Find the distance between (1, 5) and (5, 2).
Make \begin{align*}A(1, 5)\end{align*} and \begin{align*}B(5, 2)\end{align*}. Plug into the distance formula.
\begin{align*}d & = \sqrt{(1 - 5)^2 + (5 - 2)^2}\\ & = \sqrt{(-4)^2 + (3)^2}\\ & = \sqrt{16 + 9 } = \sqrt{25} = 5\end{align*}
You might recall that the distance formula was presented as \begin{align*}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\end{align*}, with the first and second points switched. It does not matter which point is first as long as \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are both first in each parenthesis. In Example 7, we could have switched \begin{align*}A\end{align*} and \begin{align*}B\end{align*} and would still get the same answer.
\begin{align*}d & = \sqrt{(5 - 1)^2 + (2 - 5)^2}\\ & = \sqrt{(4)^2 + (-3)^2}\\ & = \sqrt{16 + 9} = \sqrt{25} = 5\end{align*}
Also, just like the lengths of the sides of a triangle, distances are always positive.
Example C
Determine if the following triangles are acute, right or obtuse.
a)
b)
Set the shorter sides in each triangle equal to \begin{align*} a \end{align*} and \begin{align*} b \end{align*} and the longest side equal to \begin{align*}c\end{align*}.
a) \begin{align*}6^2 + (3 \sqrt{5})^2 \ &? \ 8^2\\ 36 + 45 \ &? \ 64\\ 81 &> 64\end{align*}
The triangle is acute.
b) \begin{align*}15^2 + 14^2 \ &? \ 21^2\\ 225 + 196 \ &? \ 441\\ 421 &< 441\end{align*}
The triangle is obtuse.
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter8ApplicationsofthePythagoreanTheoremB
Concept Problem Revisited
To find the length and width of a 52” HDTV, plug in the ratios and 52 into the Pythagorean Theorem. We know that the sides are going to be a multiple of 16 and 9, which we will call \begin{align*}n\end{align*}.
\begin{align*}(16n)^2 + (9n)^2 & = 52^2\\ 256n^2 + 81n^2 & = 2704\\ 337n^2 & = 2704\\ n^2 & = 8.024\\ n & = 2.83\end{align*}
Therefore, the dimensions of the TV are \begin{align*}16(2.83'')\end{align*} by \begin{align*}9(2.833'')\end{align*}, or \begin{align*}45.3''\end{align*} by \begin{align*}25.5''\end{align*}. If the diagonal is \begin{align*}y''\end{align*} long, it would be \begin{align*}n \sqrt{337}''\end{align*} long. The extended ratio is \begin{align*}9 : 16 : \sqrt{337}\end{align*}.
Vocabulary
The two shorter sides of a right triangle (the sides that form the right angle) are the legs and the longer side (the side opposite the right angle) is the hypotenuse. The Pythagorean Theorem states that \begin{align*}a^2+b^2=c^2\end{align*}, where the legs are “\begin{align*}a\end{align*}” and “\begin{align*}b\end{align*}” and the hypotenuse is “\begin{align*}c\end{align*}”. Acute triangles are triangles where all angles are less than \begin{align*}90^\circ\end{align*}. Right triangles are triangles with one \begin{align*}90^\circ\end{align*} angle. Obtuse triangles are triangles with one angle that is greater than \begin{align*}90^\circ\end{align*}.
Guided Practice
1. Graph \begin{align*}A(-4, 1), B(3, 8)\end{align*}, and \begin{align*}C(9, 6)\end{align*}. Determine if \begin{align*}\triangle ABC\end{align*} is acute, obtuse, or right.
2. Do the lengths 7, 8, 9 make a triangle that is right, acute, or obtuse? 3. Do the lengths 14, 48, 50 make a triangle that is right, acute, or obtuse?
Answers:
1. This looks like an obtuse triangle, but we need proof to draw the correct conclusion. Use the distance formula to find the length of each side.
\begin{align*}AB & = \sqrt{(-4-3)^2 + (1 - 8)^2} = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2}\\ BC & = \sqrt{(3 - 9)^2 + (8 - 6)^2} = \sqrt{36 + 4 } = \sqrt{40} = 2 \sqrt{10}\\ AC & = \sqrt{(-4-9)^2 + (1-6)^2} = \sqrt{169 + 25} = \sqrt{194}\end{align*}
Now, let’s plug these lengths into the Pythagorean Theorem.
\begin{align*}\left ( \sqrt{98} \right )^2 + \left ( \sqrt{40} \right )^2 & \ ? \ \left ( \sqrt{194} \right )^2\\ 98 + 40 & \ ? \ 194\\ 138 & < 194\end{align*}
\begin{align*}\triangle ABC\end{align*} is an obtuse triangle.
2. Acute because \begin{align*}7^2 + 8^2>9^2\end{align*}.
3. Right because \begin{align*}14^2+48^2=50^2\end{align*}
Practice
Find the area of each triangle below. Round your answers to the nearest tenth.
Find the length between each pair of points.
- (-1, 6) and (7, 2)
- (10, -3) and (-12, -6)
- (1, 3) and (-8, 16)
- What are the length and width of a \begin{align*}42''\end{align*} HDTV? Round your answer to the nearest tenth.
- Standard definition TVs have a length and width ratio of 4:3. What are the length and width of a \begin{align*}42''\end{align*} Standard definition TV? Round your answer to the nearest tenth.
- Challenge An equilateral triangle is an isosceles triangle. If all the sides of an equilateral triangle are \begin{align*}s\end{align*}, find the area, using the technique learned in this section. Leave your answer in simplest radical form.
- Find the area of an equilateral triangle with sides of length 8.
- The two shorter sides of a triangle are 9 and 12.
- What would be the length of the third side to make the triangle a right triangle?
- What is a possible length of the third side to make the triangle acute?
- What is a possible length of the third side to make the triangle obtuse?
- The two longer sides of a triangle are 24 and 25.
- What would be the length of the third side to make the triangle a right triangle?
- What is a possible length of the third side to make the triangle acute?
- What is a possible length of the third side to make the triangle obtuse?
- The lengths of the sides of a triangle are \begin{align*}8x, 15x,\end{align*} and \begin{align*}17x\end{align*}. Determine if the triangle is acute, right, or obtuse.
Determine if the following triangles are acute, right or obtuse.
- 5, 12, 15
- 13, 84, 85
- 20, 20, 24
- 35, 40, 51
- 39, 80, 89
- 20, 21, 38
- 48, 55, 76
Graph each set of points and determine if \begin{align*}\triangle ABC\end{align*} is acute, right, or obtuse.
- \begin{align*}A(3, -5), B(-5, -8), C(-2, 7)\end{align*}
- \begin{align*}A(5, 3), B(2, -7), C(-1, 5)\end{align*}
- Writing Explain the two different ways you can show that a triangle in the coordinate plane is a right triangle.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
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Show More |
Term | Definition |
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Distance Formula | The distance between two points and can be defined as . |
Obtuse Triangle | An obtuse triangle is a triangle with one angle that is greater than 90 degrees. |
Pythagorean Theorem | The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by , where and are legs of the triangle and is the hypotenuse of the triangle. |
Vertex | A vertex is a point of intersection of the lines or rays that form an angle. |
Image Attributions
Here you'll learn several applications of the Pythagorean Theorem.